Flo's Restaurant Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1519    Accepted Submission(s): 465 Problem Description Sick and tired of pushing paper in the dreary bleary-eyed world of finan…

Problem Description Princess Alisha invites her friends to come to her birthday party. Each of her friends will bring a gift of some value v, and all of them will come at a different time. Because the lobby is not large enough, Alisha can only let a…

[poj2424]Flo's Restaurant Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2960   Accepted: 929 Description Sick and tired of pushing paper in the dreary bleary-eyed world of finance, Flo ditched her desk job and built her own restaurant.…

Flo's Restaurant Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 932    Accepted Submission(s): 285 Problem DescriptionSick and tired of pushing paper in the dreary bleary-eyed world of finance,…

HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011 亚洲北京赛区网络赛题目) Eliminate Witches! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 863    Accepted Submission(s): 342 Problem Description Kaname Mado…

有点麻烦.. /*hdu6136[模拟+优先队列] 2017多校8*/ #include <bits/stdc++.h> using namespace std; typedef long long LL; struct frac { LL p, q; frac(LL P, LL Q) { p = P / __gcd(P, Q); q = Q / __gcd(P, Q); } frac() {p = , q = ;} bool operator < (const frac& a)…

题意:邀请k个朋友,每个朋友带有礼物价值不一,m次开门,每次开门让一定人数p(如果门外人数少于p,全都进去)进来,当最后所有人都到了还会再开一次门,让还没进来的人进来,每次都是礼物价值高的人先进.最后给出q个数,表示要输出第ni个进来的人的名字. 析:其实这就是一个模拟题,很容易知道是优先队列模拟,不能set,会超时,反正我是超时了,然后就一步步的模拟就好了,注意它给的时间可能不是按顺序,要排序,在比赛时我就没排序,一直WA.... 代码如下: #include <cstdio> #inclu…

TIANKENG's restaurant Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 0    Accepted Submission(s): 0 Problem Description TIANKENG manages a restaurant after graduating from ZCMU, and tens of t…

题目链接:https://vjudge.net/contest/184966#problem/A 题目大意: 走迷宫.从某个方向进入某点,优先走左或是右.如果左右都走不通,再考虑向前.绝对不能往后走,即使没走过. 解题思路: 重点是定义vis数组,每个点的四个方向都走过,这个点才算vis完了.……不过我还是没想明白,为什么能够想到要这样定义vis数组,应该题目做多了就知道了吧. 其它要注意的点就是如何实现优先左右转弯的功能. #include<stdio.h> #include<stri…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1789 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem DescriptionIgnatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Eve…